POJ 1087 A Plug for UNIX 會議室插座問題 構圖+最大流

POJ 1087 A Plug for UNIX 會議室插座問題 構圖+最大流

題目鏈接：POJ 1087 A Plug for UNIX

A Plug for UNIX Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 13809 Accepted: 4623

Description

You are in charge of setting up the press room for the inaugural meeting of the United Nations Internet eXecutive (UNIX), which has an international mandate to make the free flow of information and ideas on the Internet as cumbersome and bureaucratic as possible.
Since the room was designed to accommodate reporters and journalists from around the world, it is equipped with electrical receptacles to suit the different shapes of plugs and voltages used by appliances in all of the countries that existed when the room was built. Unfortunately, the room was built many years ago when reporters used very few electric and electronic devices and is equipped with only one receptacle of each type. These days, like everyone else, reporters require many such devices to do their jobs: laptops, cell phones, tape recorders, pagers, coffee pots, microwave ovens, blow dryers, curling
irons, tooth brushes, etc. Naturally, many of these devices can operate on batteries, but since the meeting is likely to be long and tedious, you want to be able to plug in as many as you can.
Before the meeting begins, you gather up all the devices that the reporters would like to use, and attempt to set them up. You notice that some of the devices use plugs for which there is no receptacle. You wonder if these devices are from countries that didn't exist when the room was built. For some receptacles, there are several devices that use the corresponding plug. For other receptacles, there are no devices that use the corresponding plug.
In order to try to solve the problem you visit a nearby parts supply store. The store sells adapters that allow one type of plug to be used in a different type of outlet. Moreover, adapters are allowed to be plugged into other adapters. The store does not have adapters for all possible combinations of plugs and receptacles, but there is essentially an unlimited supply of the ones they do have.

Input

The input will consist of one case. The first line contains a single positive integer n (1 <= n <= 100) indicating the number of receptacles in the room. The next n lines list the receptacle types found in the room. Each receptacle type consists of a string of at most 24 alphanumeric characters. The next line contains a single positive integer m (1 <= m <= 100) indicating the number of devices you would like to plug in. Each of the next m lines lists the name of a device followed by the type of plug it uses (which is identical to the type of receptacle it requires). A device name is a string of at most 24 alphanumeric
characters. No two devices will have exactly the same name. The plug type is separated from the device name by a space. The next line contains a single positive integer k (1 <= k <= 100) indicating the number of different varieties of adapters that are available. Each of the next k lines describes a variety of adapter, giving the type of receptacle provided by the adapter, followed by a space, followed by the type of plug.

Output

A line containing a single non-negative integer indicating the smallest number of devices that cannot be plugged in.

Sample Input

4 
A 
B 
C 
D 
5 
laptop B 
phone C 
pager B 
clock B 
comb X 
3 
B X 
X A 
X D 

Sample Output

1

Source

East Central North America 1999
分析：

發現基本構圖好了，網絡流的題目就很好解了。

（1）以0為源點，1為匯點，其他的插座還有設備都作為中間點

（2）會議室提供n個插座，從源點到每個插座連一條邊，容量為1

（3）會議室有m個設備，從每個設備到匯點連一條邊，容量為1

（4）每個設備使用一個插座，從相應插座到設備連一條邊，容量為1

（5）有k中轉接器，從插頭到轉接器提供插座類型連一條邊，即前者可以轉化為後者，容量為無窮，因為可以串聯。

（6）求從源點到匯點最大流，及最多使用設備數目maxflow，最後結果為m-maxflow。

代碼;

Dinic：

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define maxn 1010
#define INF 0x3f3f3f3f

struct Edge
{
    int from, to, cap;
};

vector EG;
vector G[maxn];
int n, s, t, d[maxn], cur[maxn], mp[maxn][maxn];
char name[maxn][30];
int cnt;
void addEdge(int from, int to, int cap)
{
    EG.push_back((Edge){from, to, cap});
    EG.push_back((Edge){to, from, 0});
    int x = EG.size();
    G[from].push_back(x-2);
    G[to].push_back(x-1);
}

bool bfs()
{
    memset(d, -1, sizeof(d));
    queue q;
    q.push(s);
    d[s] = 0;
    while(!q.empty())
    {
        int x = q.front();
        q.pop();
        for(int i = 0; i < G[x].size(); i++)
        {
            Edge& e = EG[G[x][i]];
            if(d[e.to] == -1 && e.cap > 0)
            {
                d[e.to] = d[x]+1;
                q.push(e.to);
            }
        }
    }
    return (d[t]!=-1);
}

int dfs(int x, int a)
{
    if(x == t || a == 0) return a;
    int flow = 0, f;
    for(int& i = cur[x]; i < G[x].size(); i++)
    {
        Edge& e = EG[G[x][i]];
        if(d[x]+1 == d[e.to] && (f = dfs(e.to, min(a, e.cap))) > 0)
        {
            e.cap -= f;
            EG[G[x][i]^1].cap += f;
            flow += f;
            a -= f;
            if(a == 0) break;
        }
    }
    return flow;
}
int Dinic()
{
    int ans = 0;
    while(bfs())
    {
        memset(cur, 0, sizeof(cur));
        ans += dfs(s, INF);
    }
    EG.clear();
    for(int i = 0; i < n; ++i)
        G[i].clear();
    return ans;
}

int Find(char* str)
{
    int i;
    for(i = 2; i < cnt; ++i)
        if(strcmp(name[i], str) == 0)
            return i;
    strcpy(name[i], str);
    cnt++;
    return i;
}
int main()
{
    //freopen("poj_1087.txt", "r", stdin);
    int m, k;
    char str1[30], str2[30];
    while(~scanf("%d", &n)) {
        s = 0, t = 1;
        cnt = 2;    //源點和匯點占兩個
        for(int i = 0; i < n; i++) {
            scanf("%s", str1);
            strcpy(name[cnt], str1);    //插座不會從父，直接插入
            cnt++;
            addEdge(0, i+2, 1);     //建邊，源點向每個插座連一條1的邊
        }
        scanf("%d", &m);
        for(int i = 0; i < m; i++) {
            scanf("%s%s", str1, str2);
            strcpy(name[cnt], str1);    //設備也不會重復。直接插入
            cnt++;
            int u = Find(str2);         //扎裡插座可能重復，要查找
            addEdge(u, cnt-1, 1);   //建邊，插座向設備連一條容量為1的邊
            addEdge(cnt-1, 1, 1);       //建邊，設備到匯點連一條邊，容量為1
        }
        scanf("%d", &k);
        for(int i = 0; i < k; i++) {
            scanf("%s%s", str1, str2);
            int u = Find(str1);
            int v = Find(str2);
            addEdge(v, u, INF);     //建邊，後者到前者容量為無窮
        }
        n = cnt;
        int ans = Dinic();
        printf("%d\n", m-ans);
    }
    return 0;
}

ISAP：

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define maxn 1010
#define INF 0x3f3f3f3f

struct Edge
{
    int from, to, cap, flow;
};

char name[maxn][30];
vector EG;
vector G[maxn];
int n, s, t, d[maxn], cur[maxn], p[maxn], num[maxn], mp[maxn][maxn];
bool vis[maxn];
int cnt;
void addEdge(int from, int to, int cap)
{
    EG.push_back((Edge){from, to, cap, 0});
    EG.push_back((Edge){to, from, 0, 0});
    int x = EG.size();
    G[from].push_back(x-2);
    G[to].push_back(x-1);
}

void bfs()
{
    memset(vis, false, sizeof(vis));
    queue q;
    vis[t] = true;
    d[t] = 0;
    q.push(t);
    while(!q.empty()) {
        int x = q.front();
        q.pop();
        for(int i = 0; i < G[x].size(); i++) {
            Edge& e = EG[G[x][i]^1];
            if(!vis[e.from] && e.cap > e.flow) {
                vis[e.from] = true;
                d[e.from] = d[x]+1;
                q.push(e.from);
            }
        }
    }
}

int augment()
{
    int x = t, a = INF;
    while(x != s) {
        Edge& e = EG[p[x]];
        a = min(a, e.cap-e.flow);
        x = EG[p[x]].from;
    }
    x = t;
    while(x != s) {
        EG[p[x]].flow += a;
        EG[p[x]^1].flow -= a;
        x = EG[p[x]].from;
    }
    return a;
}
int ISAP()
{
    int ans =0;
    bfs();
    memset(num, 0, sizeof(num));
    for(int i = 0; i < n; i++)
        num[d[i]]++;
    int x = s;
    memset(cur, 0, sizeof(cur));
    while(d[s] < n) {
        if(x == t) {
            ans += augment();
            x = s;
        }
        bool flag = false;
        for(int i = cur[x]; i < G[x].size(); i++) {
            Edge& e = EG[G[x][i]];
            if(e.cap > e.flow && d[x] == d[e.to]+1) {
                flag = true;
                p[e.to] = G[x][i];
                cur[x] = i;
                x = e.to;
                break;
            }
        }
        if(!flag) {
            int m = n-1;
            for(int i = 0; i < G[x].size(); i++) {
                Edge& e = EG[G[x][i]];
                if(e.cap > e.flow)
                    m = min(m, d[e.to]);
            }
            if(--num[d[x]] == 0) break;
            num[d[x] = m+1]++;
            cur[x] = 0;
            if(x != s)
                x = EG[p[x]].from;
        }
    }
    EG.clear();
    for(int i = 0; i < n; ++i)
        G[i].clear();
    return ans;
}
int Find(char* str)
{
    int i;
    for(i = 2; i < cnt; ++i)
        if(strcmp(name[i], str) == 0)
            return i;
    strcpy(name[i], str);
    cnt++;
    return i;
}
int main()
{
    //freopen("poj_1087.txt", "r", stdin);
    int m, k;
    char str1[30], str2[30];
    while(~scanf("%d", &n)) {
        s = 0, t = 1;
        cnt = 2;
        for(int i = 0; i < n; i++) {
            scanf("%s", str1);
            strcpy(name[cnt], str1);
            cnt++;
            addEdge(0, i+2, 1);
        }
        scanf("%d", &m);
        for(int i = 0; i < m; i++) {
            scanf("%s%s", str1, str2);
            strcpy(name[cnt], str1);
            cnt++;
            int u = Find(str2);
            addEdge(u, cnt-1, 1);
            addEdge(cnt-1, 1, 1);
        }
        scanf("%d", &k);
        for(int i = 0; i < k; i++) {
            scanf("%s%s", str1, str2);
            int u = Find(str1);
            int v = Find(str2);
            addEdge(v, u, INF);
        }
        n = cnt;
        int ans = ISAP();
        printf("%d\n", m-ans);
    }
    return 0;
}

不知為何，總感覺我的Dinic比ISAP要快，基本每次都是，不應該啊。